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LostInParadise's avatar

What do you think of this high school level math problem?

Asked by LostInParadise (32183points) March 6th, 2020

I just thought of this, and I like the way that a little intuition goes a long way in setting up the simple algebra needed to solve it.

You have a starting point of 0 and an end point of 1.

Move the starting point to the middle, giving start and end points of ½ and 1.

Move the end point half way, giving a starting point of ½ and end point ¾.

Now repeat the above two steps, alternately moving the starting and end points half way. Where do you eventually end up?

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9 Answers

gorillapaws's avatar

It’s an asymptote. Are you looking for that answer or the value that each point approaches but never reaches? The ambiguity is frustrating. It’s explained well otherwise.

Pinguidchance's avatar

Two thirds of me likes the question, the other third not so much.

LostInParadise's avatar

It seems that you have found the answer,

@gorillapaws , Yes I am looking for the limit point.

Here is a hint. Each time we change the start and end values, we get a new problem that is a scaled down version of the original. Express the solution of the new problem in terms of the original. This gives an equation with one variable that can be easily solved for.

gorillapaws's avatar

@LostInParadise “Yes I am looking for the limit point.”

Then this sentence is inaccurate:

“Where do you eventually end up?”

This process will repeat forever, so the question itself presumes a falsehood and is therefore inaccurate. I would rephrase it.

LostInParadise's avatar

Your point is well taken, but I was speaking informally. I am going to keep it as it is.

The way I solved it was to realize that when the start and end are ½ and ¾, it is essentially the same problem. The length is now ¼ with offset of ½, and if x is the limit point with respect to the original interval, then ½ + ¼ x should take us to the same place.
We get x = ½ + ¼ x.
¾x = ½
x = 4/3(½) = ⅔

There is another way of looking at this. The equation f(x) =½ + ¼ x maps the interval from 0 to 1 onto the interval from ½ to ¾. The equation x = ½ + ¼ x gives the value that maps to itself. f(⅔) = ⅔. Intuitively (the algebra would be straight forward but tedious), ⅔ will keep mapping to itself on each iteration. The size of the interval keeps shrinking, so it must be that ⅔ is the limit.

zenvelo's avatar

This is one of Zeno’s paradoxes, Achilles and the Tortoise.

gorillapaws's avatar

@LostInParadise “I am going to keep it as it is.”

I guess I misunderstood the point of the question. I was thinking you wanted us to vette a test question for your students (i.e. is it clear, fair, etc), not that your intent was to challenge the Fluther community to solve the problem. I wasn’t suggesting you edit the question on the site—just trying to help improve the clarity for your students. Apologies for the misunderstanding.

LostInParadise's avatar

I am not a teacher, but am an academic at heart. I have a Web site covering topics in high school math, and try to find things that I could add. I am not sure if I want to include this problem, but if I do, the term limit will be used.

dxs's avatar

I think it will be tricky for a high school student unless they’re hinted at some useful tools beforehand.

I solved this problem by finding a limit of the sequence of starting/end points. For instance, 1, ¾, 5/8, 11/32, 85/128, ... was the sequence of end points. The general expressions for both sequences were of the form ⅔ + [something that went to 0].

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